To prove the converse of the theorem we follow the method
in [1]}. Assume \(L^1(K)\)  to be \(\alpha \) -left
amenable and consider \(\alpha \) -left \(L^1(K)\) -module \(L^\infty (K)\) 
with the module multiplications
\(f\cdot \varphi =\widehat{f}(\alpha )\varphi \) 
and \(\varphi \cdot f=f\ast \varphi \) ,
for every \(f\in L^1(K)\)  and
\(\varphi \in L^\infty (K)\) .
Since \(\alpha \cdot f=f\ast \alpha =\widehat{f}(\alpha )\alpha \) ,
\({C}\alpha \)  is a closed \(L^1(K)\) -submodule of \(L^\infty (K)\) .
Hence, \(L^\infty (K)=X\oplus {C}\alpha \)  where \(X\)  is also
a closed \(L^1(K)\) -submodule of \(L^\infty (K).\)  Choose
\(\nu \in L^\infty (K)^\ast \)  such that \(\nu (\alpha )=1\) , and define
\(\delta :L^1(K)\rightarrow L^\infty (K)^\ast ,\)  \(\delta (f)=f\cdot \nu -\nu \cdot f\) ,
for \(f\in L^1(K).\) 
Then
\(\delta (f)(\alpha )&=f\cdot \nu (\alpha )-\nu \cdot f(\alpha )\\ &=\nu (\alpha \cdot f)-\nu (f\cdot \alpha )\\ &=\nu (f\ast \alpha )- \widehat{f}(\alpha )\nu (\alpha )\\&=\widehat{f}(\alpha )\nu (\alpha )-\widehat{f}(\alpha )\nu (\alpha )=0,\hspace{28.45274pt}\mbox{ for every } f\in L^1(K).\) 
